5 Marks
Qs.1. We Care Air needs to make a decision about Flight 105. There are currently 3 seats
reserved for last minute customers, but the airline does not know if anyone will buy them. If
they release the seats now, they know they will be able to sell them for $250 each. Last
minute customers must pay $475 per seat. The decision must be made now, and any number
of seats may be released. We Care Air has the following probability distribution to help them:
Number of last minute customers requesting seats 0 1 2 3
Probability 0.45 0.30 0.15 0.10
The Company also counts a $150 loss of goodwill for every last-minute customer who turned
away.
a) How much revenue will be generated by releasing all 3 seats now?
b) What is the company’s expected net revenue (revenue less loss of goodwill) if 3 seats
are releases now?
c) What is the company’s expected net revenue if 2 seats are released now?
How many seats should be released to maximize expected revenue?
3 Marks
Qs.2. An international mutual funds reported strong earnings in 2007. The population of
international mutual funds earned a mean return of 14.52% in 2007. Assume that the returns
for international mutual funds were distributed as a normal random variable, with a mean of
14.57 and a standard deviation of 20. If you selected a random sample of 10 funds from this
population, what is the probability that the sample would have mean return
a. Less than 0- that is, a loss?
b. Between 0 and 20?
c. Greater than 10?
2.5 Marks
Qs.3. The unemployment rate is 4.1%. Assume that 100 employable people are randomly
selected.
a. What is the expected number of people who are unemployable?
b. What is the variance and standard deviation of the number of people who are
unemployable?
a. E(x) = np = 100(.041) = 4.1
b. Var(x) = np(1 - p) = 100(.041)(.959) = 3.93
3.93 1.98
σ
= =
2.5 Marks
Qs.4. Owner of the Aurora Restaurant is considering purchasing new furniture. To help
him decide on the amount he can afford to invest in table and chairs, he wishes to determine
the average revenue per customer. The check for 9 randomly sampled customers had an
average of $18.30 and a standard deviation of $3.60. Construct a 95% confidence interval for
the size of the average check per customer.
Docsity.com
